Câu hỏi:
Cho \(\int\limits_0^{\frac{\pi }{2}} {f\left( x \right){\rm{d}}x} = 5\). Khi đó \(I = \int\limits_0^{\frac{\pi }{2}} {\left[ {f\left( x \right) + 2\sin x} \right]{\rm{d}}x} \) bằng
A. \(I = 7\).
B. \(I = 5 + \frac{\pi }{2}\).
C. \(I = 3\).
D. \(I = 5 + \pi \).
GY:
Ta có: \(I = \int\limits_0^{\frac{\pi }{2}} {\left[ {f\left( x \right)\, + 2\sin x} \right]\,{\rm{d}}x = \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\,{\rm{d}}x\,{\rm{ + 2}}\int\limits_0^{\frac{\pi }{2}} {\sin x\,{\rm{d}}x} } } \)
\( = \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)} \,{\rm{d}}x\,\, – \left. {2\cos x} \right|_0^{\frac{\pi }{2}} = 5 – 2\left( {0 – 1} \right) = 7\)
Vậy: \(I = 7\).
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